5
$\begingroup$

I have a large amount of copper sulfate available and I am interested in obtaining copper chloride from it. What is the easiest way to do so?

I have considered mixing an aqueous solution of copper sulfate with sodium chloride, but I think that it would be difficult to separate the resulting sodium sulfate from the copper chloride. My other idea is to first react the copper sulfate with sodium carbonate and then react the resulting copper carbonate with hydrochloric acid, which would produce copper chloride (the other products would be $\ce{CO2}$ gas and water). I am trying to find a cost effective method for performing this reaction.

$\endgroup$
4
$\begingroup$

You cannot obtain $\ce{CuCl2}$ mixing $\ce{CuSO4}$ and $\ce{NaCl}$ because both are very soluble, the second strategy seems good because $\ce{CuCO3}$ is insoluble in water, and you can react the new salt with $\ce{HCl}$.

Good luck.

$\endgroup$
4
$\begingroup$

Here's a much more direct method: just add a stoichiometric amount of calcium chloride to a concentrated aqueous solution of copper sulfate:

$$\ce{CuSO4(aq) + CaCl2(aq) -> CaSO4(s) + CuCl2(aq)}$$

If the copper sulfate you have is in the anhydrous form (white powder), then add 695.3 mg of pure $\ce{CaCl2}$ per 1000 mg of pure $\ce{CuSO4}$. If your copper sulfate is in the form of the pentahydrate ($\ce{CuSO4.5H2O}$, striking blue crystals), then add 444.5 mg of pure $\ce{CaCl2}$ per 1000 mg of pure $\ce{CuSO4.5H2O}$.

The low solubility of calcium sulfate (~0.2 g per 100 g of water at 20°C) is what drives the reaction forwards. This double displacement reaction does not happen if you use sodium chloride, because sodium sulfate is highly soluble. Simply filter the solid out and obtain a relatively pure solution of copper(II) chloride, with small amounts of calcium and sulfate contamination. Drying the solution completely will net $\ce{CuCl2}$ crystals.

$\endgroup$
0
$\begingroup$

I would guess you can directly mix copper sulfate solution with concentrated $\ce{HCl}$ to yield dissolved $\ce{CuCl2}$ and $\ce{H2SO4}$. This is of course in competition with (re-)formation of $\ce{CuSO4}$ but I would be interested in seeing how excess $\ce{HCl}$ would affect yield.

Also -- I made $\ce{CuCl2}$ by dissolving copper metal in concentrated $\ce{HCl}$, adding concentrated $\ce{H2O2}$ (slowly) and then evaporating off the water solvent. The dehydrated form is a brown powder and moderately irratiting. Easily soluble in water and has a blue/green solution color. -- So some method of reducing the copper sulfate to copper-metal (e.g. electrolysis), and then this dissolution in $\ce{HCl/H2O2}$ should do it.

Acid fumes deem a fume hood necessary, but decomposition of $\ce{H2O2}$ just gives off water and oxygen gas.

Along the lines of your precipitating idea, using barium chloride or lead chloride to form the corresponding insoluble sulfate would potentially be a faster, one step process, although perhaps more expensive. After, you decant the remaining $\ce{CuCl2}$ and evaporate solvent.

$\endgroup$
  • 2
    $\begingroup$ Adding conc. $\ce{HCl}$ to $\ce{CuSO4}$ solution will give a yellow solution containing $\ce{[CuCl4]^{2-}}$ ions. It definitely won't produce $\ce{CuCl2}$ as all the ions will stay in solution. $\endgroup$ – bon Jul 31 '15 at 15:08
  • $\begingroup$ I thought it may form a more complex ion, thanks. By the way --how does one write subscripts without italicizing (with LaTex)? I'd like to contribute here more often. @bon $\endgroup$ – khaverim Jul 31 '15 at 15:13
  • $\begingroup$ @bon Nevermind, I can see it from your edits, ty $\endgroup$ – khaverim Jul 31 '15 at 15:15
  • 1
    $\begingroup$ See here for information on formatting. $\endgroup$ – bon Jul 31 '15 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.